document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=826407>Question 826407</A>: <I><SPAN STYLE=\"word-break: break-all;\"> It is true in general, that the perpendicular bisectors of the sides of a triangle are concurrent and that the point of intersection is the centre of the circle through all three vertices. Prove this result by placing the vertices at A(2a,0), B(-2a,0) and C(2b,2c) and proceeding as follows:\r<BR>\n" );
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document.write( "a)  Find the gradients of AB,BC and CA, and hence find the equations of the three perpendicular bisectors. \r<BR>\n" );
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document.write( "While I do know how to do the rest of the steps (that's why I didn't write it down), I seem to be having trouble with the first step (written above). Would someone be able to help me with this particular question?<BR>\n" );
document.write( "Thanks!!  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #498077 by </B><A HREF=/tutors/aboutme.mpl?userid=KMST><B>KMST(5328)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=KMST><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=KMST><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=498077\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For each line, the gradient (we call it slope) is the differences in y-coordinates divided by the difference in x-coordinates.<BR>\n" );
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document.write( "Segment AB is part of the horizontal line <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=0 ALIGN=MIDDLE  ALT=\"y=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > , the x-axis.<BR>\n" );
document.write( "Its midpoint is (0,0) , the origin.<BR>\n" );
document.write( "The perpendicular bisector to AB is the line perpendicular to the x-axis through the origin. That is the line <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=0 ALIGN=MIDDLE  ALT=\"x=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > , the y-axis.<BR>\n" );
document.write( "All that is so by design (clever choice of coordinates, placing the triangle so it would be so).<BR>\n" );
document.write( " <BR>\n" );
document.write( "The other two perpendicular bisectors are not so easy to find.<BR>\n" );
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document.write( "The gradient/slope of AC is<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5BAC%5D=%28y%5BC%5D-y%5BA%5D%29%2F%28x%5BC%5D-x%5BA%5D%29=%282c-0%29%2F%282b-2a%29=2c%2F%282%28b-a%29%29=c%2F%28b-a%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "A perpendicular line would have a gradient/slope of<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%29%2F%28c%2F%28b-a%29%29=-%28b-a%29%2Fc=%28a-b%29%2Fc ALIGN=MIDDLE  ALT=\"%28-1%29%2F%28c%2F%28b-a%29%29=-%28b-a%29%2Fc=%28a-b%29%2Fc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The midpoint of AC has<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=%28x%5BC%5D%2Bx%5BA%5D%29%2F2=%282b%2B2a%29%2F2=2%28b%2Ba%29%2F2=b%2Ba ALIGN=MIDDLE  ALT=\"x=%28x%5BC%5D%2Bx%5BA%5D%29%2F2=%282b%2B2a%29%2F2=2%28b%2Ba%29%2F2=b%2Ba\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%28y%5BC%5D%2By%5BA%5D%29%2F2=%282c%2B0%29%2F2=2c%2F2=c ALIGN=MIDDLE  ALT=\"y=%28y%5BC%5D%2By%5BA%5D%29%2F2=%282c%2B0%29%2F2=2c%2F2=c\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "With that we can write the equation for the perpendicular bisector to AC in point-slope form as<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-c=%28%28a-b%29%2Fc%29%28x-%28b%2Ba%29%29 ALIGN=MIDDLE  ALT=\"y-c=%28%28a-b%29%2Fc%29%28x-%28b%2Ba%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The intersection of that line and the line <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=0 ALIGN=MIDDLE  ALT=\"x=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > has <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=0 ALIGN=MIDDLE  ALT=\"x=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > (of course) and<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-c=%28%28a-b%29%2Fc%29%280-%28b%2Ba%29%29 ALIGN=MIDDLE  ALT=\"y-c=%28%28a-b%29%2Fc%29%280-%28b%2Ba%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-c=-%28a-b%29%28b%2Ba%29%2Fc ALIGN=MIDDLE  ALT=\"y-c=-%28a-b%29%28b%2Ba%29%2Fc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%28b%5E2-a%5E2%29%2Fc%2Bc ALIGN=MIDDLE  ALT=\"y=%28b%5E2-a%5E2%29%2Fc%2Bc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%28c%5E2%2Bb%5E2-a%5E2%29%2Fc ALIGN=MIDDLE  ALT=\"y=%28c%5E2%2Bb%5E2-a%5E2%29%2Fc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( " <BR>\n" );
document.write( "Similarly,<BR>\n" );
document.write( "the midpoint of BC has<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=%28x%5BC%5D%2Bx%5BB%5D%29%2F2=%282b-2a%29%2F2=2%28b-a%29%2F2=b-a ALIGN=MIDDLE  ALT=\"x=%28x%5BC%5D%2Bx%5BB%5D%29%2F2=%282b-2a%29%2F2=2%28b-a%29%2F2=b-a\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%28y%5BC%5D%2By%5BB%5D%29%2F2=%282c%2B0%29%2F2=2c%2F2=c ALIGN=MIDDLE  ALT=\"y=%28y%5BC%5D%2By%5BB%5D%29%2F2=%282c%2B0%29%2F2=2c%2F2=c\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "The gradient of BC is<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5BBC%5D=%28y%5BC%5D-y%5BB%5D%29%2F%28x%5BC%5D-x%5BB%5D%29=%282c-0%29%2F%282b%2B2a%29=2c%2F%282%28b%2Ba%29%29=c%2F%28b%2Ba%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "A perpendicular line would have a gradient/slope of<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%29%2F%28c%2F%28b%2Ba%29%29=-%28b%2Ba%29%2Fc=-%28a%2Bb%29%2Fc ALIGN=MIDDLE  ALT=\"%28-1%29%2F%28c%2F%28b%2Ba%29%29=-%28b%2Ba%29%2Fc=-%28a%2Bb%29%2Fc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "With that we can write the equation for the perpendicular bisector to AC in point-slope form as<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-c=%28-%28a%2Bb%29%2Fc%29%28x-%28b-a%29%29 ALIGN=MIDDLE  ALT=\"y-c=%28-%28a%2Bb%29%2Fc%29%28x-%28b-a%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "So the intersection of that line and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=0 ALIGN=MIDDLE  ALT=\"x=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > has <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=0 ALIGN=MIDDLE  ALT=\"x=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > (of course) and<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-c=%28-%28a%2Bb%29%2Fc%29%280-%28b-a%29%29 ALIGN=MIDDLE  ALT=\"y-c=%28-%28a%2Bb%29%2Fc%29%280-%28b-a%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-c=%28-%28a%2Bb%29%2Fc%29%28a-b%29 ALIGN=MIDDLE  ALT=\"y-c=%28-%28a%2Bb%29%2Fc%29%28a-b%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-c=-%28a%2Bb%29%28a-b%29%2Fc ALIGN=MIDDLE  ALT=\"y-c=-%28a%2Bb%29%28a-b%29%2Fc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-c=-%28a%5E2-b%5E2%29%2Fc ALIGN=MIDDLE  ALT=\"y-c=-%28a%5E2-b%5E2%29%2Fc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y-c=%28b%5E2-a%5E2%29%2Fc ALIGN=MIDDLE  ALT=\"y-c=%28b%5E2-a%5E2%29%2Fc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%28b%5E2-a%5E2%29%2Fc%2Bc ALIGN=MIDDLE  ALT=\"y=%28b%5E2-a%5E2%29%2Fc%2Bc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%28c%5E2%2Bb%5E2-a%5E2%29%2Fc ALIGN=MIDDLE  ALT=\"y=%28c%5E2%2Bb%5E2-a%5E2%29%2Fc\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( " <BR>\n" );
document.write( "So all 3 perpendicular bisectors pass through the point with<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=0 ALIGN=MIDDLE  ALT=\"x=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y=%28c%5E2%2Bb%5E2-a%5E2%29%2Fc ALIGN=MIDDLE  ALT=\"y=%28c%5E2%2Bb%5E2-a%5E2%29%2Fc\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "Since all points on the perpendicular bisector of a segment are equidistant from the ends of the segment, the point above should be at the same distance from A, B, and C, so a circle centered at that point, with that equal distance as its radius would pass through A, B, and C.<BR>\n" );
document.write( "Its radius can be calculated as the distance from that point to A(2a,0)<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28%280-2a%29%5E2%2B%28%28c%5E2%2Bb%5E2-a%5E2%29%2Fc-0%29%5E2%29=sqrt%284a%5E2%2B%28c%5E2%2Bb%5E2-a%5E2%29%5E2%2Fc%5E2%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > </SPAN>\n" );
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